Hi, this is partially a question from physics, but I think mathematicians will also know the answer:
Why is $\frac{1}{j \omega C} = \frac{-j}{\omega C}$?
You may not care for the answer but here it is:
it is the same reason some insist that $\dfrac{1}{\sqrt 2}$ be written as $\dfrac{\sqrt 2}{2}$.
Both are holdovers from pre-calculator days when calculations were made easier by rationalizing the denominator. Dividing by two is easy. Same can be said of $\bf{i}$. Moreover $\dfrac{1}{\bf{i}}=\dfrac{-\bf{i}}{1}$ is in fact true.
It's counter-intuitive that $\frac1{\mathbb j}=-\mathbb j$, but to see it clearly:
$$\frac1{\mathbb j} = \frac{\mathbb j}{\mathbb j} \cdot \frac1{\mathbb j} = \frac{\mathbb j}{\mathbb j \cdot \mathbb j} = \frac{\mathbb j}{\mathbb j^2} = \frac{\mathbb j}{-1}=-{\mathbb j}$$